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In Short Put Spreads, why buy an A put, and sell a B put? If $A < p < B$, you can be assigned to your B put, while your A put is worthless. Kevin Ott diagrams this below, but I added A, B.

Instead, why not buy a B put, and sell an A put? Then if $A < p < B$, you can't be assigned, and your B put is ITM. So you profit more!

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This is going to get ugly :->)

There are 6 basic synthetic positions relating to combinations of put options, call options and their underlying stock (the Synthetic Triangle):

  1. Synthetic Long Stock = Long Call + Short Put
  2. Synthetic Short Stock = Short Call + Long Put
  3. Synthetic Long Call = Long Stock + Long Put
  4. Synthetic Short Call = Short Stock + Short Put
  5. Synthetic Short Put = Long Stock + Short Call
  6. Synthetic Long Put = Short Stock + Long Call

These are all variations of S + P - C = 0

Suppose the strike prices of your short put spread are -$50p and +45p then by using #s 5 and 6 from above:

-50p = +STK - 50c

+45p = -STK + 45c

Add the two equations together and you get:

-50p + 45p = 45c - 50c

Therefore, selling the 50/45 put spread is equivalent to buying to 45/50 call spread and NOT as you suggested, just reversing the buy and sell of the respective puts in the spread.

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Instead, why not buy a B put, and sell an A put?

That is a different strategy, a bear/debit put spread, equivalent to a bear/credit call spread, which is shown in your other question. If you are comparing the value of the position for a given $p$, you must take into account that a debit spread costs cash up front, whereas a credit spread pays cash up front. The profit/loss plot includes this cash component. You cannot compare the value of the options alone.

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